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void setup(){ Serial.begin(9600); } void loop(){ static unsigned long count=0; static float sum=0; float ave=0; int volt=analogRead(A0); float voltage = map(volt,0,1023,0,2500) / 100.0; Serial.print("Voltage: "); Serial.print(voltage,2); sum=sum+voltage; count++; ave=sum/count; Serial.print(" Average: "); ...


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Arduino's power connector jack's tip is positive while the sleeve is negative but the more important thing to know is what will you be using in powering Arduino? Please provide more information about your setup and possibly a schematic. This will help us understand more what you're trying to accomplish.


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It doesn't output 0v because it's near impossible. 0.003v is in fact very low. The LOW output, according to the datasheet, can be as high as 0.8v (when powered by 5v) and still be considered valid. LOW does not mean 0v. It means "below the voltage threshold that is considered a LOW".


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What about the amperage? (I guess that the max amperage of the USB power supply is divided between the board and sensors/actuators connected). How does it compare with the VCC output of, let say, an Arduino UNO? The 5V pin is connected directly to the 5V of the USB. Whatever your USB port can provide the 5V can provide, minus the current used by the ESP8266....


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If you scroll down on that page you link to you can see the power circuit. It uses an XC6206 voltage regulator which has a maximum input voltage of 7V and a dropout voltage that could be as high as 680mV. So you can feed between about 4V and 7V in to the VIN pad and it will step it down to 3.3V. Anything below 4V and you will go "out of regulation" ...


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For this task you should use an extra power supply not running trough the Arduino. For this kind of setup I use PCA9865 16-channel servo dirver board cost around 5$, needs two pins only and can run all servos (strong 5V power supply provided => Amperage) simultaniously. For a max of 8 servos a 5V/2-2.5A power supply (wall wart) should be sufficient. See this ...


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I suspect the USB 5v supply is rather noisy and sags slightly with load. A hall effect sensor measures TINY signals, so it's probably quite sensitive to changes in supply voltage. You might try adding a capacitor to the line, but you'd be better off either feeding 7.5-9-VDC into the barrel connector and using the on-board voltage regulator on the Arduino to ...


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It sounds like your +5v supply is only barely adequate and sags as you draw even marginally more current. But even given that, your AREF is the TL431's voltage which 1) won't match the supply voltage (slightly redrawn schematic shows why), and 2) the sensor voltage could possibly exceed the AREF which is out of spec for the Arduino's A/D. A more stable ...


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TL;DR: You really need to know what components you have and you need to find datasheets. In the datasheet, you find what voltage the component requires. Don't care too much about amperes in the beginning. If you get the voltage right, each component will just draw as much current as it needs. An Arduino Uno operates from 7 to 12 Volts, so 7.4 V is ok. "...


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"Arduino Pro Mini hooked up to the 5V power supply via the RAW and GND pins, but the supply is a regulated 5v wall supply. " That is the problem. RAW pin expects more like 6V to 6.5V for the regulator to output 5V. If you are supplying 5V, connect to the 5V, or VCC and GND pins. If you are connecting to USB via an FTDI Basic or equivalent, and the 5V ...


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Your code is not very long, but I already see a few things that I would change. int analogPin = 0; [...] pinMode(A4, INPUT); These usages of pins are inconsistent. I personally prefer settings at the beginning of the file, so I'd expect int analogPin = 0; int whateverPin = A4; // I don't have a good name yet While you have pinMode(A4, INPUT); there's ...


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This is exactly why the typical LiPO battery shield includes a boost converter. I'm not sure where you saw the unusually high (3.5V) low-voltage limit, but that isn't typical of most converters.


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Actually it is two separate questions and I answer your first question of reading multiple analog values using one ADC and the answer is Yes you can use this approach but you should give enough time for each channel to charge internal ADC capacitor when you change channel .so if you use more independent ADC you get more time for each channel and result is ...


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The maximum voltage at the Vin pin is 12V, so using your battery is fine! Considering the current consumption: The arduino itself will usually source less than 200mA, but your whole setup will probably sink more than that.


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